Vector Magnitudes given only direction-help!

1. The problem statement, all variables and given/known data
You leave your campsite to go get supplies at a store. You travel 230M at an angle 31 degrees south of east to get there(I'll call this vector A for simplicity's sake). On the way back to camp, you go a distance B at 43 degrees north of west and a distance C at 60 degrees south of west. Find the distances B and C.

2. Relevant equations
Not sure. I've tried the law of sins: sin(a)/A=sin(b)/B=sin(c)/C. I can't think of anyway to use the dot product or cross product to solve this problem, though that's what the section is supposedly about.

3. The attempt at a solution
I used the law of sins first. Initially, I used the angles provided and got a wrong answer, as I didn't manipulate the measures of the angles. Later, I tried making the X-axis go along A and adjusting the angle between A and B to 12 degrees and the angle between B and C as 103 degrees. This also gave a wrong answer of 213.95. (I had the sin(103)/230=sin(65)/B)). I got the 65 degree based on triangles having 180 degrees.

I've also attempted setting up a system of equations for the components of B and C, as I know that the X and Y components of B and C must add up to be the opposite of the X and Y components of vector A, but I hit a wall there as well.

I haven't attempted finding the distance of C until I am sure B is correct.

I have no idea why the dot or cross product is pertinent in the case, if it is at all.

I have a feeling I am either missing a very simple concept, making a math error in the angles I am using, or just making the problem harder than it needs to be.

1. The problem statement, all variables and given/known data
You leave your campsite to go get supplies at a store. You travel 230M at an angle 31 degrees south of east to get there(I'll call this vector A for simplicity's sake). On the way back to camp, you go a distance B at 43 degrees north of west and a distance C at 60 degrees south of west. Find the distances B and C.

2. Relevant equations
Not sure. I've tried the law of sins: sin(a)/A=sin(b)/B=sin(c)/C. I can't think of anyway to use the dot product or cross product to solve this problem, though that's what the section is supposedly about.

3. The attempt at a solution
I used the law of sins first. Initially, I used the angles provided and got a wrong answer, as I didn't manipulate the measures of the angles. Later, I tried making the X-axis go along A and adjusting the angle between A and B to 12 degrees and the angle between B and C as 103 degrees. This also gave a wrong answer of 213.95. (I had the sin(103)/230=sin(65)/B)). I got the 65 degree based on triangles having 180 degrees.

I've also attempted setting up a system of equations for the components of B and C, as I know that the X and Y components of B and C must add up to be the opposite of the X and Y components of vector A, but I hit a wall there as well.

I haven't attempted finding the distance of C until I am sure B is correct.

I have no idea why the dot or cross product is pertinent in the case, if it is at all.

I have a feeling I am either missing a very simple concept, making a math error in the angles I am using, or just making the problem harder than it needs to be.

Please point me in the right direction!!

Thanks for any help, it's really appreciated.

Lets reserve comment on the laws of sin. But that said you will need to solve for both of your unknowns with what you do know.

Since you end up back at the same place - camp - then the x,y components of all the vectors must add to 0. This should allow you to construct 2 equations - one for the sum of x components in terms of angles and unknowns and likewise for all the y components. Then you can solve for B and C.

For instance 240*Cos31 - B*Cos43 - C*Cos60 = 0
Develop the equation for Y and then you can calculate for B and C.

Thanks a lot! I was making this question way too hard... I forgot how to solve simple systems of equations. I kept getting 0=0 because when I tried it before, I substituted the variable I solved for into the same equation rather than whichever I hadn't solved initially.